7.35 moles of oxygen. Initially, for each mole of H₂CO₃, there are 3 moles of oxygen, as derived from the acid's formula. For 2.45 moles of the compound stated in the problem, which is carbonic acid, we calculate: If 1 mole of H₂CO₃ corresponds to 3 moles of oxygen, then for 2.45 moles of H₂CO₃, we have X moles of oxygen. Thus, X = (3 × 2.45) / 1 = 7.35 moles of oxygen.
The result is 4.16 L.
Based on the provided information, we calculated the following:
Molarity = 0.225 M,
Quantity of KI = 0.935 moles,
To find Volume: Molarity = moles/Volume, hence Volume = moles/Molarity.
Thus, Volume = 0.935/0.225, giving Volume = 4.16 L.
Consequently, 4.16 L of KI is required.
Mass of Ag = 1.67 g Mass of Cl = 2.21 g Heat released = 1.96 kJ Explanation: To find the enthalpy of formation of AgCl(s), we have the reaction: 2Ag(s) + Cl2(g) → 2AgCl(s). First, we need to calculate the moles of Ag and Cl from their masses. The atomic mass of Ag is 108 g/mol, thus moles of Ag = 1.67/108 = 0.0155 moles. The atomic mass of Cl is 35 g/mol, therefore moles of Cl = 2.21/35 = 0.0631 moles. Since Ag's moles are much less than Cl's, silver is the limiting reagent. Based on the stoichiometry of the reaction, the number of moles of AgCl produced equals 0.0155 moles. The enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol. Hence, the enthalpy of formation is 126.5 kJ/mol.
Answer: The change in enthalpy will be -13.
Explanation:-
Endothermic reactions absorb heat, while exothermic reactions release heat. In the case of an endothermic reaction, the change in enthalpy is represented as positive, whereas for an exothermic reaction, it is negative.
When one mole of A combines with one mole of B to form three moles of C
So the stoichiometric ratio being halved also results in the enthalpy for the reaction being halved.
Thus, for this reaction:
The resulting change in enthalpy is -13.
